NCERT Exemplar Class 12 Maths Solutions Relations and Function...

Correct option: (D) Reflexive, transitive but not symmetric.

Concept used. A relation R on N is reflexive if nRn holds for every n∈N; symmetric if nRm ⇒ mRn; transitive if nRm and mRr ⇒ nRr. Here nRm ⇔ nm, i.e. n divides m.

1. Reflexive: every n∈N divides itself, so nRn holds. Yes.
2. Symmetric: take n=3, m=6. Then 36 so 3R6, but 63, so 6R 3. Hence R is not symmetric.
3. Transitive: if nm and mr, then nr (basic divisibility), so nRmmRr⇒ nRr.

R is reflexive and transitive but not symmetric; option (D).

Correct option: (B) symmetric.

Concept used. `` perpendicular to m'' is a geometric condition on the angle between two lines being 90^∘.

1. Reflexive? A line is never perpendicular to itself (the angle is 0^∘), so R is false. Not reflexive.
2. Symmetric? If ⊥ m, then certainly m⊥ –- perpendicularity is mutual. Yes.
3. Transitive? Take horizontal, m vertical, n horizontal. Then ⊥ m and m⊥ n, but ∥ n, so ⊥ n. Not transitive.

Only symmetry holds; option (B).

Correct option: (B) injective.

Concept used. f is one–one/injective if f(n₁)=f(n₂)⇒ n₁=n₂; f is onto/surjective if every y in the codomain is hit by some n in the domain.

1. Injective: 2n₁+3 = 2n₂+3 ⇒ 2n₁ = 2n₂ ⇒ n₁=n₂. So f is one–one.
2. Surjective on N? The smallest value of f is f(1)=5. Numbers 1,2,3,4 in the codomain are never hit, so f is not onto. The values of f are exactly 5,7,9,…, a proper subset of N.

f is injective but not surjective; option (B).

Correct option: (C) 24.

Concept used. The number of injective (one–one) functions from a set of m elements into a set of n elements (with n≥ m) is the permutation ⁿP_m = n!(n-m)!. Each of the m domain elements must go to a distinct codomain element; for the first there are n choices, for the second n-1, and so on.

1. Here |A|=m=3, |B|=n=4.
2. Number of injections = ⁴P₃ = 4× 3× 2 = 24.

24 injective mappings; option (C).

Correct option: (C) sin x².

Concept used. The composition of two functions is (f∘ g)(x) = f(g(x)): feed g first, then apply f to the result.

1. Start inside: g(x)=x².
2. Apply f to g(x): (f∘ g)(x) = f(g(x)) = f(x²) = sin(x²).

(f∘ g)(x)=sin x²; option (C).

Correct option: (A) x+43.

Concept used. If f is invertible, set y=f(x), solve for x in terms of y, then rename: f^-1(y) = x, and finally swap y→ x to get f^-1(x).

1. Let y = 3x-4.
2. Solve for x:    3x = y+4 ⇒ x = y+43.
3. So f^-1(y) = y+43, which gives f^-1(x) = x+43.
4. Check: f(f^-1(x)) = 3·x+43-4 = (x+4)-4 = x.

f^-1(x)=x+43; option (A).

Correct option: (D) 5.

Concept used. Equivalence relations on a set A are in one-to-one correspondence with partitions of A. The number of partitions of an n-element set is the Bell number B_n. For n=3, B₃ = 5.

1. Partition with 1 block: 1,2,3 –- gives the universal relation A× A.
2. Partitions with 2 blocks: 1,2,3, 2,1,3, 3,1,2 –- three of them.
3. Partition with 3 blocks (singletons): 1,2,3 –- the identity (diagonal) relation.
4. Total = 1 + 3 + 1 = 5.

5 equivalence relations; option (D).

Correct option: (B) (x-5)^1/3.

Concept used. For a bijective f, invert by solving y=f(x) for x. Here f(x)=x³+5 is strictly increasing on R, so it is a bijection R→R and f^-1 exists.

1. Let y=x³+5.
2. x³=y-5 ⇒ x=(y-5)^1/3 (real cube root, defined for all y∈R).
3. Therefore f^-1(x)=(x-5)^1/3.
4. Check: f(f^-1(x)) = ((x-5)^1/3)³+5 = (x-5)+5 = x.

f^-1(x)=(x-5)^1/3; option (B).

Correct option: (B) [1,∞).

Concept used. Complete the square to find the minimum value of a quadratic, then read off the range over the given domain.

1. Complete the square: x²-4x+5 = (x-2)²+1.
2. Vertex of the parabola at x=2, minimum value f(2)=1.
3. Domain is [2,∞), where the parabola is strictly increasing.
4. As x runs from 2 to ∞, f(x) runs from 1 to ∞. Range = [1,∞).

Range =[1,∞); option (B).

Correct option: (A) reflexive.

Concept used. For any real x, x-x+√2=√2, which is irrational. So reflexivity is immediate. We test the other two properties by counter-examples.

1. Reflexive: xRx ⇔ x-x+√2=√2 is irrational –- true for every x∈R.
2. Symmetric: take x=√2, y=0. Then x-y+√2 = 2√2 (irrational), so √2 R 0. But y-x+√2 = -√2+√2=0 (rational), so 0R √2. Not symmetric.
3. Transitive: take x=√2, y=1, z=2√2. x-y+√2=2√2-1 (irrational), y-z+√2=1-√2 (irrational), but x-z+√2=0 (rational). Not transitive.

Only reflexivity holds; option (A).

Required pairs: (b,b), (c,c), (a,c).

Concept used. R is reflexive on A iff (x,x)∈ R for every x∈ A. R is transitive iff (x,y),(y,z)∈ R ⇒ (x,z)∈ R.

1. For reflexivity on A=a,b,c, we need (a,a),(b,b),(c,c)∈ R. Already (a,a)∈ R, so add (b,b) and (c,c).
2. For transitivity check the existing pairs: (a,b)∈ R and (b,c)∈ R, so we must have (a,c)∈ R. Add (a,c).
3. Now check no new transitive obligation arises from the added pairs: (a,c),(c,c)⇒(a,c) already present; (a,a),(a,b)⇒(a,b) already present; etc. All triples close.

Add exactly 3 pairs: (b,b),(c,c),(a,c).

Domain: D = [-5, 5].

Concept used. A real-valued square root √u requires the radicand u≥ 0. So the domain is the set of x for which 25-x²≥ 0.

1. Set up the inequality: 25-x²≥ 0.
2. Rearrange: x²≤ 25.
3. Solve: |x|≤ 5, i.e. -5≤ x≤ 5.
4. Both endpoints are included since √0=0 is defined.

D = [-5, 5].

Answer: (g∘ f)(x) = 4x²+4x-1.

Concept used. (g∘ f)(x) = g(f(x)): substitute the entire expression for f(x) into g.

1. Compute f(x)=2x+1.
2. Substitute into g: g(f(x)) = g(2x+1) = (2x+1)² - 2.
3. Expand: (2x+1)² = 4x²+4x+1.
4. So (g∘ f)(x) = 4x²+4x+1-2 = 4x²+4x-1.

(g∘ f)(x) = 4x²+4x-1.

Answer: f^-1 = (b,a),(d,b),(a,c),(c,d).

Concept used. A function on a finite set is invertible iff it is a bijection (one–one and onto). Each input maps to a distinct output, and every element of A appears as an output, so f is a bijection A→ A. Its inverse f^-1 is obtained by reversing every ordered pair.

1. List f: a→ b, b→ d, c→ a, d→ c.
2. Outputs =b,d,a,c=A and all distinct: f is a bijection.
3. Reverse each pair: b→ a, d→ b, a→ c, c→ d.
4. Write f^-1 as a set of ordered pairs.

f^-1=(b,a),(d,b),(a,c),(c,d).

Claim. f(x)=cos x on R→R is neither injective nor surjective.

Concept used. f is one–one iff distinct inputs give distinct outputs; f is onto iff every y∈ R is hit. Recall the range of cosine: [-1,1], and that cos is 2π-periodic.

1. Not one–one. Take x₁=0, x₂=2π with x₁≠ x₂. Then cos 0 = cos 2π = 1, so f(x₁)=f(x₂) without x₁=x₂. Injectivity fails.
2. Not onto. For any x∈R, -1x≤ 1. Hence the range of f is [-1,1]R. The value y=2∈R is never attained, so surjectivity fails.

f is neither one–one nor onto.

Claim. R is an equivalence relation on Z (this is congruence modulo n, written a≡ bn).

Concept used. ``n divides k'' means k = n for some ∈Z. We must verify all three of reflexivity, symmetry, and transitivity.

1. Reflexive. For every a∈Z, a-a=0=n· 0, which is divisible by n. So aRa holds.
2. Symmetric. Suppose aRb, i.e. a-b = n for some ∈Z. Then b-a = -n= n(-), which is again a multiple of n. So bRa.
3. Transitive. Suppose aRb and bRc: a-b=n₁ and b-c=n₂. Adding, a-c=(a-b)+(b-c)=n₁+n₂= n(₁+₂), divisible by n. So aRc.
4. All three properties hold, so R is an equivalence relation.

R is an equivalence relation on Z –- this is congruence modulo n.

Claim. f is one–one and onto, hence bijective.

Concept used. To show one–one, assume f(x₁)=f(x₂) and deduce x₁=x₂. To show onto, take a generic y∈ B and produce a preimage x∈ A with f(x)=y.

1. One–one. Suppose x₁-2x₁-3 = x₂-2x₂-3. Cross-multiply: (x₁-2)(x₂-3) = (x₂-2)(x₁-3).
2. Expand both sides: x₁x₂ - 3x₁ - 2x₂ + 6 = x₁x₂ - 3x₂ - 2x₁ + 6.
3. Cancel common terms: -3x₁ - 2x₂ = -3x₂ - 2x₁, hence -x₁ = -x₂, i.e. x₁ = x₂. So f is one–one.
4. Onto. Let y∈ B = R-1. We must find x∈ A with f(x)=y.
5. Set x-2x-3=y ⇒ x-2 = y(x-3) = yx-3y ⇒ x-yx = 2-3y ⇒ x = 2-3y1-y, which is well-defined because y≠ 1.
6. Check x≠ 3: if x=3 then 2-3y1-y=3 ⇒ 2-3y=3-3y ⇒ 2=3, contradiction. So x∈ A.
7. Hence every y∈ B has a preimage in A: f is onto.

f is bijective with f^-1(y)=2-3y1-y.

Claim. f is neither injective nor surjective on R.

Concept used. For one–one, equate f(x₁)=f(x₂) and check if x₁=x₂ is forced. For onto, examine the range.

1. Set x₁x₁²+1 = x₂x₂²+1. Cross-multiply: x₁(x₂²+1) = x₂(x₁²+1).
2. Expand: x₁x₂²+x₁ = x₂x₁²+x₂ ⇒ x₁x₂(x₂-x₁) = x₂-x₁ ⇒ (x₂-x₁)(x₁x₂ - 1) = 0.
3. Either x₁=x₂ or x₁x₂ = 1.
4. Take x₁=2, x₂=12: distinct, but f(2)=25 and f(12) = 1/21/4+1 = 1/25/4= 25. Equal images, distinct inputs –- not one–one.
5. For onto, test y=1: xx²+1=1 ⇒ x = x²+1 ⇒ x²-x+1 = 0, whose discriminant is 1-4=-3<0. No real solution, so 1 is not in the range.
6. Hence f is not onto R.

f is neither one–one nor onto.

Answer. (i) (f∘ g)(x)=4x²-6x+1;    (ii) (g∘ f)(x)=2x²+6x-1.

Concept used. (f∘ g)(x)=f(g(x)): replace every x in f(x) with the full expression for g(x) and simplify. Likewise (g∘ f)(x)=g(f(x)).

1. (i) f∘ g. f(g(x)) = f(2x-3) = (2x-3)²+3(2x-3)+1.
2. Expand (2x-3)²=4x²-12x+9.
3. Expand 3(2x-3)=6x-9.
4. Add: 4x²-12x+9+6x-9+1 = 4x²-6x+1.
5. (ii) g∘ f. g(f(x)) = 2f(x)-3 = 2(x²+3x+1)-3 = 2x²+6x+2-3 = 2x²+6x-1.

f∘ g = 4x²-6x+1;   g∘ f = 2x²+6x-1.

Claim. R is reflexive, symmetric, and transitive; the class of (2,5) is [(2,5)] = (1,4),(2,5),(3,6),(4,7),(5,8),(6,9).

Concept used. The condition a+d = b+c is equivalent to a-b = c-d: the relation says ``two ordered pairs have the same first-minus-second.'' This is exactly the kind of ``equal-invariant'' condition that defines an equivalence relation.

1. Reflexive. For (a,b)∈ A× A: a+b=b+a trivially, so (a,b)R(a,b).
2. Symmetric. If (a,b)R(c,d) then a+d=b+c, i.e. c+b = d+a, which is the condition (c,d)R(a,b). So R is symmetric.
3. Transitive. Suppose (a,b)R(c,d) and (c,d)R(e,f). Then a+d=b+c and c+f=d+e. Add the two equations: a+d+c+f=b+c+d+e⇒ a+f=b+e, which is (a,b)R(e,f).
4. All three hold, so R is an equivalence relation.
5. Compute [(2,5)]. (c,d)∈[(2,5)] ⇔ 2+d=5+c ⇔ d-c=3. We need c,d∈1,…,9 with d=c+3: (1,4),(2,5),(3,6),(4,7),(5,8),(6,9).

[(2,5)]=(1,4),(2,5),(3,6),(4,7),(5,8),(6,9) –- 6 ordered pairs.